Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
B. isolation joints
C. control joints
D. construction joints
If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provide Control joints. The correct answer would be C.
Control joints are used to prevent random cracks from forming in large concrete slabs caused by shrinkage. These joints are placed at strategic locations in the slab to allow for the concrete to expand and contract without cracking. Expansion joints, on the other hand, are used to separate concrete from other structures or materials, and isolation joints are used to separate different sections of concrete.
Construction joints are used to connect two different pours of concrete. Therefore, the best option for preventing random cracks caused by shrinkage would be to use control joints.
Learn more about Control joints:
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Cut that photo by
1. Left click your mouse on the photo
2. Click cut
Then enter the file where you want to transfer and press
1. ctrl+v
Answer:
you can go to your file and then select the phpto and hold on a little bit and choose the delete option
Answer:
(iv) second law of thermodynamics
Explanation:
The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero
Answer:
The correct answer is: the following factors are needed to properly consider while selecting a brake or clutch:
-Engagement
-Friction
-Electromagnetic
-Mechanical
-Actuation
-Electric
-Fluid power
-Self-actuation
-Key concepts
-Application notes
-Selection criteria
Explanation:
Clutches and brakes are important devices in many rotating drive systems, it is very important to guarantee the security and the proper function of them accomplishing a high quality parameters in those factors.
Answer:
Explanation:
Given that,
Average fiber diameter is 0.01mm
d = 0.01mm = 1 × 10^-5m
The average fiber length is 2.5mm
L = 2.5mm = 0.0025m
Volume of the fraction of fibers is 0.40
Vf = 0.40
Fiber matrix bond strengths is 75MPa
τ = 75 MPa
The fraction strength of the fibers is 3500 Mpa
σf = 3500 MPa
The matrix street fiber is 8 MPa
σm = 8 MPa
We need to find the critical fiber length and compare it to original fiber length
Ic = σf•d / 2τ
Ic = 3500 × 0.01 / 75 × 2
Ic = 0.233 mm
Since the critical fiber length of 0.233 mm is much less than the provided length of the fiber (2.5mm) , so we can use the following equation to find the longitudinal tensile strength
σcd = σf•Vf(1—Ic / 2L) + σm(1—Vf)
σcd = 3500×0.4[1—0.233/(2 × 2.5)] + 8(1—0.4)
σcd = 1400(1—0.0467) + (8 × 0.6)
σcd = 1334.67 + 4.8
σcd = 1339.47 MPa
The longitudinal tensile strength of the aligned glass fiber-epoxy matrix composite is 1339.47 MPa
Answer:
Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.
Explanation:
The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.
Mathematically we can write,
where,
E(t) is the modulus of elasticity at any temperature 'T'
is the modulus of elasticity at absolute zero.
is the mean melting point of the material
Hence we can see that with increasing temperature modulus of elasticity decreases.
In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.